Exercises — Physical pendulum — compound pendulum

$T=2\pi \sqrt{\frac{I}{mgd}},I=I_{cm}+m{d}^2=m{k}^2+m{d}^2,L_{\text{eq}}=\frac{I}{md}=\frac{k^2}{d}+d$ Yahan $I$ moment of inertia hai pivot ke baare mein, $d$ pivot-se-CM ki distance hai, $m$ mass hai, aur $k$ CM ke baare mein radius of gyration hai. Hum $g=9.8{\text{m/s}}^2$ lete hain jab tak koi problem kuch aur na kahe.

Sab kuch already "pivot ke baare mein" hai, toh bas $T=2\pi \sqrt{I/(mgd)}$ mein substitute karo: $T=2\pi \sqrt{\frac{0.30}{2.0\cdot 9.8\cdot 0.25}}=2\pi \sqrt{\frac{0.30}{4.9}}=2\pi \sqrt{0.06122}$ $T=2\pi (0.2474)\approx \boxed{1.55\text{s}}$ Humne kya kiya: diye gaye numbers seedhe daal diye — koi theorem ki zaroorat nahi kyunki $I$ already pivot ke baare mein bata diya gaya tha.

Parent note se, ek end par pivoted rod ka $L_{\text{eq}}=\frac{2}{3}\ell$ hota hai (kyunki $I_{end}=\frac{1}{3}m{\ell }^2$ aur $d=\ell /2$, toh $L_{\text{eq}}=I/(md)=\frac{1}{3}{\ell }^2/\frac{1}{2}\ell =\frac{2}{3}\ell$). $L_{\text{eq}}=\frac{2}{3}(1.2)=\boxed{0.80\text{m}}$ Iska matlab: yeh poori rod bilkul waisi swing karti hai jaise 0.80 m ki string par latki ek ball.

Step 1 — pivot ke baare mein $I$ banao. Disk ke liye, $I_{cm}=\frac{1}{2}m{R}^2$. Pivot rim par hai, CM se $d=R$ door, toh Parallel axis theorem se: $I=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2=1.5(0.80)(0.15{)}^2=0.0270{\text{kg⋅m}}^2$ Step 2 — $d$ identify karo. $d=R=0.15\text{m}$. Step 3 — period. $T=2\pi \sqrt{\frac{\frac{3}{2}m{R}^2}{mgR}}=2\pi \sqrt{\frac{3R}{2g}}=2\pi \sqrt{\frac{3(0.15)}{2(9.8)}}=2\pi \sqrt{0.02296}$ $T=2\pi (0.1515)\approx \boxed{0.952\text{s}}$ Dhyaan do mass cancel ho gaya — gravity pendulums mein hamesha aisa hota hai. Neeche disk-at-rim geometry dekho.

Step 1 — pivot ke baare mein $I$. Ek ring ka poora mass radius $R$ par hota hai, toh $I_{cm}=m{R}^2$. Pivot edge par hai, $d=R$: $I=m{R}^2+m{R}^2=2m{R}^2$ Step 2 — $d=R$. Step 3 — $T=2\pi \sqrt{\frac{2m{R}^2}{mgR}}=2\pi \sqrt{\frac{2R}{g}}=2\pi \sqrt{\frac{2(0.20)}{9.8}}=2\pi \sqrt{0.04082}$ $T=2\pi (0.2020)\approx \boxed{1.27\text{s}}$ Sanity check: $L_{\text{eq}}=I/(md)=2m{R}^2/(mR)=2R=0.40\text{m}$ — ek hooped ring aise swing karti hai jaise diameter ke barabar length ka simple pendulum ho. Zabardast.

(a) Ek rod ka radius of gyration apne centre ke baare mein $I_{cm}=\frac{1}{12}m{\ell }^2=m{k}^2$ se milta hai, toh $k=\frac{\ell }{\sqrt{12}}=\frac{1.0}{3.464}=0.2887\text{m}.$ Period minimum hota hai jab $d=k$ (parent note §4). Toh $\boxed{d=k=0.289\text{m}}$ (b) Minimum par, $L_{\text{eq}}=2k$, toh $T_{\min }=2\pi \sqrt{\frac{2k}{g}}=2\pi \sqrt{\frac{2(0.2887)}{9.8}}=2\pi \sqrt{0.05892}=2\pi (0.2427)\approx \boxed{1.525\text{s}}$ Neeche figure mein dikhaya gaya hai ki $T$ is minimum tak kaise girta hai aur phir wapas chadh jaata hai.

End-pivot period (rod formula $T=2\pi \sqrt{2\ell /(3g)}$ se): $T_{end}=2\pi \sqrt{\frac{2(1.0)}{3(9.8)}}=2\pi \sqrt{0.06803}=2\pi (0.2608)\approx 1.639\text{s}$ Ratio: $\frac{T_{end}}{T_{\min }}=\frac{1.639}{1.525}\approx \boxed{1.075}$ Toh end se swing karna sirf lagbhag 7.5% slower hai fastest possible pivot se — minimum $d=k$ ke paas broad aur shallow hai, isliye exact pivot point ka itna fark nahi padta.

(a) $L_{\text{eq}}=\frac{2}{3}\ell =0.667\text{m}$ (rod at end). (b) Center of oscillation pivot se $L_{\text{eq}}=0.667\text{m}$ ki distance par hai, CM se guzarne wali line par. Kyunki CM $0.50$ m par hai, center of oscillation CM se $0.667-0.500=0.167\text{m}$ aage hai. (c) Wahan pivot karo. Ab nayi pivot-to-CM distance $d^{\prime }=0.167\text{m}$ hai. Radius of gyration $k$ unchanged hai ($k=0.2887$ m). Toh $L_{\text{eq}}^{\prime }=\frac{k^2}{d^{\prime }}+d^{\prime }=\frac{0.08333}{0.1667}+0.1667=0.5000+0.1667=0.667\text{m}$ Same $L_{\text{eq}}$! Isliye same period. Yeh reversibility exactly wahi hai jo Kater's pendulum $I$ ya $m$ jaane bina $g$ measure karne ke liye use karta hai.

Step 1 — pivot ke baare mein total moment of inertia. Dono contributions add karo (ek hi axis ke baare mein moments of inertia add hote hain): $I=\underset{\text{rod, end}}{\underbrace{\frac{1}{3}m{\ell }^2}}+\underset{\text{bob at end}}{\underbrace{M{\ell }^2}}=\frac{1}{3}(0.50)(1{)}^2+(0.50)(1{)}^2=0.1667+0.5000=0.6667{\text{kg⋅m}}^2$ Step 2 — combined CM locate karo. Rod ka CM $0.50$ m par, bob $1.0$ m par: $d=\frac{m(0.50)+M(1.0)}{m+M}=\frac{0.25+0.50}{1.0}=0.75\text{m}$ Step 3 — total mass aur period. $m_{tot}=1.0\text{kg}$: $T=2\pi \sqrt{\frac{I}{m_{tot}gd}}=2\pi \sqrt{\frac{0.6667}{1.0\cdot 9.8\cdot 0.75}}=2\pi \sqrt{0.09070}=2\pi (0.3012)\approx \boxed{1.892\text{s}}$ Humne kya synthesise kiya: $I$ aur $d$ dono do objects combine karke aaye — yahan ek single table value use nahi ho sakti.

$L_{\text{eq}}=k^2/d+d$ aur $T=2\pi \sqrt{L_{\text{eq}}/g}$ use karo: $L_{\text{eq}}=\frac{(0.2887{)}^2}{0.010}+0.010=\frac{0.08333}{0.010}+0.010=8.333+0.010=8.343\text{m}$ $T=2\pi \sqrt{\frac{8.343}{9.8}}=2\pi \sqrt{0.8513}=2\pi (0.9227)\approx \boxed{5.80\text{s}}$ Trend: $d\to 0$ hone par, $k^2/d$ term dominate karta hai aur $L_{\text{eq}}\to \infty$, toh $T\to \infty$. Physically restoring torque $mgd\sin \theta$ vanish ho jaata hai jabki $I$ finite rehta hai — rod, apne balance point ke almost pass pivoted, return karna hi nahi chahti, toh aur aur slowly swing karti hai. Yeh L3 figure mein U-curve ki divergent left branch hai.

(a) $T=2\pi \sqrt{L_{\text{eq}}/g}$ se, $L_{\text{eq}}$ ke liye solve karo: $L_{\text{eq}}=\frac{g{T}^2}{4{\pi }^2}=\frac{9.8(0.85{)}^2}{4{\pi }^2}=\frac{9.8(0.7225)}{39.48}=\frac{7.0805}{39.48}\approx \boxed{0.1794\text{m}}$ (b) Kyunki $L_{\text{eq}}=k^2/d+d$, $k$ ke liye rearrange karo: $k^2=d(L_{\text{eq}}-d)=0.12(0.1794-0.12)=0.12(0.05935)=0.007122$ $k=\sqrt{0.007122}\approx \boxed{0.0844\text{m}}$ Humne kya master kiya: poori chain ulti chalana — stopwatch reading se body ki geometric property $k$ tak, exactly Kater's pendulum ki spirit aur experimentally Moment of inertia measure karne ki (kyunki $I_{cm}=m{k}^2$).

Ek rod ke liye end par, $T=2\pi \sqrt{2\ell /(3g)}$ — period sirf $\ell$ par depend karta hai (mass cancel ho jaata hai). Toh $\frac{T_{2\ell }}{T_{\ell }}=\sqrt{\frac{2\ell }{\ell }}=\sqrt{2}\approx \boxed{1.414}$ Lamba rod slower swing karta hai, uska period $\sqrt{2}$ se bada. Har length double karne par period $\sqrt{2}$ se multiply hota hai, $2$ se nahi — wahi square-root law jo Simple pendulum ($T\propto \sqrt{L}$) ko govern karta hai.

Page cover karo: rim par latki disk ka $L_{\text{eq}}$ kya hai? Answer ::: $L_{\text{eq}}=\frac{3}{2}R$ ($I=\frac{3}{2}m{R}^2$, $d=R$ se).